increment and decrement operator

[JATINCSE]

please told me the answer and reason for the value of a and b

#include<iostream.h>
#include<conio.h>
void main()
{int a=3;
int b = ++a + a++ + a-- + --a;
cout<<"\n enter a:"<<a;
cout<<"\n enter b:"<<b;
getch();
}

[weaknessforcats]

The answer is indeterminate.

Variables may be evaluated by the compiler in any order within a statement. If you change the value of a variable more than once inside the statement, the change may be applied in an order that's different from the one in your mind.

[JATINCSE]

please suggest me the way to handle these type of problems.
i can predict the answer of these type of statement but the way of predicting is not clear to me that what actually the compiler is doing

[weaknessforcats]

The way to handle these problems is to never change a variable more than once in any statement.

There is no way to know what the compiler is doing since different compilers behave differently.

[HSharma1729]

Output will be:
enter a:3
enter b:12

Reason:
First evaluate the values of prefix operations and then solve the problem by using using that value.
Here,
a=3
b=++a + a++ + a-- + --a
so, a=3+1-1=3 (1 prefix increment and 1 prefix decrement)
hence, b=3+3+3+3=12
a=3+1+1-1-1=3

[donbock]

@HSharma1729, those might be the answers obtained for your compiler but I assure you there are other compilers out there that will give different answers.

The C Standard explicitly says the results are undefined for an expression that changes the value of a variable more than once.

You can test your compiler to get a feel for what it does for specific instances of undefined behavior, but you can never be sure. For example, behavior frequently varies based on optimization setting. What you find out about how one compiler reacts to undefined behavior tells you nothing about how another compiler will react.

Avoid undefined behavior.